# Cauchy-Schwarz Inequality question

Find the number of ordered quadruples $$(a,b,c,d)$$ of nonnegative real numbers such that \begin{align*} a^2 + b^2 + c^2 + d^2 &= 4, \\ (a + b + c + d)(a^3 + b^3 + c^3 + d^3) &= 16. \end{align*}

I have 21 as my answer since (1,1,1,1), (0,0,0,2), and $$(0,0\sqrt{2},\sqrt{2})$$ and found all the combinations possible with (0,0,0,2) and $$(0,0\sqrt{2},\sqrt{2})$$,then adding up all the possible combinations and getting 21. However this answer is incorrect. What did I do wrong?

• $(a, b,c,d)=(0,0,2,2)$ does not satisfy the second equation. – Dietrich Burde Apr 25 at 18:13
• How is this a Cauchy-Schwarz inequality question? – kccu Apr 25 at 18:20
• He meant, that on can solve this system using the Cauchy-Schwarz inequality. – Dr. Sonnhard Graubner Apr 25 at 18:27
• @Dr.SonnhardGraubner How so? – kccu Apr 25 at 18:48
• @Dr. Sonnhard Graubner please do not assume my gender. That is rude. – sumi Apr 25 at 18:59

## 1 Answer

$$a^2 + b^2 + c^2 + d^2 = 4$$

$$(a + b + c + d)(a^3 + b^3 + c^3 + d^3) = 16$$

By the C.S. inequality $$(a + b + c + d)(a^3 + b^3 + c^3 + d^3) \ge (a^2 + b^2 + c^2 + d^2)^2$$

In this case they are equal, and the equality holds only when: $$\frac{a}{a^2} = \frac{b}{b^2} = \frac{c}{c^2} = \frac{d}{d^2}$$ $$\implies a = b = c = d = 1$$

Note that if any of $$\{a,b,c,d\}$$ are $$0$$, the ratios will not exist. We need to consider them on a case-by-case basis:

1. One of them, say $$a=0$$: In this case, $$b=c=d=\frac{2}{\sqrt{3}}$$
2. Two of them are zero, say $$a=b=0$$. Then, $$b = c = \sqrt{2}$$
3. Three of them say $$a=b=c=0$$. Then $$d=2$$

Hence the solutions are: $$\boxed{(a,b,c,d) = (1,1,1,1),\ (0, \frac{2}{\sqrt{3}},\ \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}),\ (0, 0, \sqrt{2}, \sqrt{2}),\ (0,0,0,2)}$$

Number of possibilities:

1. $$(1,1,1,1) \implies ^4C_4 = 1$$
2. $$(0, \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}) \implies ^4C_1 = 4$$
3. $$(0, 0, \sqrt{2}, \sqrt{2}) \implies ^4C_2 = 6$$
4. $$(0,0,0,2) \implies ^4C_1 = 4$$

Hence the total number of cases seems to be $$\boxed{15}$$

• That's not true, as you can see $(0,0,\sqrt{2},\sqrt{2})$ is a solution. – kccu Apr 25 at 18:52
• @kccu (-1, -1, -1, -1) is a solution as per this argument but the problem states that the numbers are non-negative. I need to check how CS works for zeros – user1952500 Apr 25 at 18:53
• If some of the terms are zero, then you need to apply C-S with fewer terms (e.g., $(a+b+c)(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2$ if $d=0$). You will find the additional solutions $(0,0,0,2)$, $(0,0,\sqrt{2},\sqrt{2})$, and $(0, \frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}})$, as well as all permutations of those. – kccu Apr 25 at 18:57
• @kccu I was just now writing down the exact same thing. Please take a look – user1952500 Apr 25 at 19:00
• Don't forget permutations of those (e.g., $(\sqrt{2},0,\sqrt{2},0)$, $(0,2,0,0)$, etc.). – kccu Apr 25 at 19:17